Question

我在Laravel有一个表格

{{ Form::label('address', trans('Address')) }} 
            @if ( $bounty->address == "NULL")
            {{ Form::text('address', Input::old('address'), array('class' => 'form-control')) }}
            @else
            {{ Form::text('address', $bounty->address, array('class' => 'form-control')) }}
            @endif

这是控制器

public function bounty($id)
{   
    $title = $this->title->byId($id);
    $bounty = array();
    $bounty = DB::table('bounties')
        ->where('title-id', $id)
        ->first();

        return View::make('Titles.EditBounty')->withTitle($title)->withBounty($bounty);

基本上，如果address字段的bounty表中已有值，我希望表单将该值设置为默认值，否则，如果应该返回空白。

但是，我一直收到此错误

尝试获取非对象的属性（查看：/home/stephenm/public_html/app/views/Titles/EditBounty.blade.php）

<?php echo Form::text('address', Input::old('address'), array('class' => 'form-control')); ?>

address字段

时没有值

Answer 1

您可以使用object_get()辅助功能：

 @if(!object_get($bounty,'address'))
   {{ Form::text('address', Input::old('address'), array('class' => 'form-control')) }}
 @else
  {{ Form::text('address', $bounty->address, array('class' => 'form-control')) }}
 @endif

甚至可以使用if默认值完全消除Input::old()条件：

{{ Form::text('address', Input::old('address', object_get($bounty,'address')), array('class' => 'form-control')) }}

Laravel：检查记录是否存在

1 个答案: